# Question about decomposition of exterior product

In their paper "New lower bounds for the border rank of matrix multiplication", Landsberg and Ottaviani make use of the fact that

$$\tag{\dagger} {\textstyle\bigwedge}^p(V\otimes W) \cong \bigoplus\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} \mathbb{S}_\lambda V \otimes \mathbb{S}_{\bar\lambda}W$$

where $\bar\lambda$ denotes the conjugate partition of $\lambda$. This isomorphism is basically Exercise 6.11 in Fulton & Harris, so there is no doubt about it. However, from what I gather, in Lemma 3.1 of the paper, they use the fact that the above isomorphism is given by the map

$$(v_1\otimes w_1)\wedge\ldots\wedge(v_p\otimes w_p) \longmapsto \sum\nolimits_{\substack{\lambda\vdash p\\\\\ell(\lambda)\le n\\\\\lambda_1\le m }} c_\lambda(v_1\otimes\ldots\otimes v_p) \otimes c_{\bar\lambda}(w_1\otimes\ldots\otimes w_p),$$

where $c_\lambda$ denotes the Young symmetrizer corresponding to the partition $\lambda$. I cannot find a proof for this. Can someone explain to me why the above map defines

a) a morphism of $\mathfrak{S}_p$-modules and
b) a bijection?

Since all vector spaces involved are of finite dimension and by $(\dagger)$, it would certainly suffice to show that it is either injective or surjective.

Also, if I misunderstood the proof of Lemma 3.1 and the isomorphism is given by another elementary rule, please tell me what it is.

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In a), you don't mean $\mathfrak S_p$ but you mean $\mathrm{GL}\left(V\right)\times \mathrm{GL}\left(W\right)$, right? – darij grinberg May 17 '12 at 16:57
I actually think it is both, and the $\mathrm{GL}(V)\times\mathrm{GL}(W)$ part seemed rather clear. Please educate me if I'm wrong. – Jesko Hüttenhain May 17 '12 at 17:35
I don't see how $\mathbb S_{\lambda} V$ (in general, not just for $\lambda$ being $\left(1,1,...,1\right)$ or $\left(p\right)$) becomes a $\mathfrak S_p$-module. We have $\mathbb S_{\lambda} V = c_{\lambda} V^{\otimes p}$, and $c_{\lambda}$ is not (in general) central in $k\mathfrak S_p$. – darij grinberg May 17 '12 at 23:35
Hm. That's actually a relief, because most of my problems came from trying to understand that. I suppose it doesn't have to carry $\mathfrak{S}_p$-module structure anyway. – Jesko Hüttenhain May 18 '12 at 5:41

The equality $\sum_n e_n(xy) = \sum_\lambda s_{\lambda}(x)s_{\bar\lambda}(y)$ does imply $(\dagger)$, but why does it mean that the isomorphism is of this form? – Jesko Hüttenhain May 17 '12 at 15:26